Divide the following complex numbers. $\dfrac{9-2i}{1-4i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${1+4i}$. $ \dfrac{9-2i}{1-4i} = \dfrac{9-2i}{1-4i} \cdot \dfrac{{1+4i}}{{1+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(9-2i) \cdot (1+4i)} {1^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(9-2i) \cdot (1+4i)} {(1)^2 - (-4i)^2} $ $ = \dfrac{(9-2i) \cdot (1+4i)} {1 + 16} $ $ = \dfrac{(9-2i) \cdot (1+4i)} {17} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({9-2i}) \cdot ({1+4i})} {17} $ $ = \dfrac{{9} \cdot {1} + {-2} \cdot {1 i} + {9} \cdot {4 i} + {-2} \cdot {4 i^2}} {17} $ $ = \dfrac{9 - 2i + 36i - 8 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{9 - 2i + 36i + 8} {17} = \dfrac{17 + 34i} {17} = 1+2i $